Urysohn's Lemma - Sketch of Proof

Sketch of Proof

For every dyadic fraction r ∈ (0,1), we are going to construct an open subset U(r) of X such that:

1. U(r) contains A and is disjoint from B for all r
2. for r < s, the closure of U(r) is contained in U(s).

Once we have these sets, we define f(x) = inf { r : x ∈ U(r) } for every x ∈ X. Using the fact that the dyadic rationals are dense, it is then not too hard to show that f is continuous and has the property f(A) ⊆ {0} and f(B) ⊆ {1}.

In order to construct the sets U(r), we actually do a little bit more: we construct sets U(r) and V(r) such that

• A ⊆ U(r) and B ⊆ V(r) for all r
• U(r) and V(r) are open and disjoint for all r
• for r < s, V(s) is contained in the complement of U(r) and the complement of V(r) is contained in U(s).

Since the complement of V(r) is closed and contains U(r), the latter condition then implies condition (2) from above.

This construction proceeds by mathematical induction. Since X is normal, we can find two disjoint open sets U(1/2) and V(1/2) which contain A and B, respectively. Now assume that n≥1 and the sets U(a/2n) and V(a/2n) have already been constructed for a = 1,...,2n-1. Since X is normal, we can find two disjoint open sets which contain the complement of V(a/2n) and the complement of U((a+1)/2n), respectively. Call these two open sets U((2a+1)/2n+1) and V((2a+1)/2n+1), and verify the above three conditions.

The Mizar project has completely formalized and automatically checked a proof of Urysohn's lemma in the URYSOHN3 file.